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Integral over scaling function

  1. Jun 22, 2011 #1

    [itex]\phi(x)[/itex] is an interpolating scaling function (also called fundamental function or Dubuc-Deslauriers function) as given on pages 155 to 158 in these lecture notes: http://pages.unibas.ch/comphys/comphys/TEACH/WS07/course.pdf [Broken]

    Why does the follwoing yield:

    [itex]\int_{-\infty}^{\infty}\phi(x) dx = 1?[/itex]

    At least, I assume this yields, because otherwise I cannot show the equality of the first equation of the exercise on page 158 of the above lecture notes:

    http://img577.imageshack.us/img577/304/capturevm.png [Broken]
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jun 23, 2011 #2


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    φ(x) seems to be the Dirac delta function. Check it out on Wikipedia if you are not familiar with it. Most authors use δ(x).
  4. Jun 24, 2011 #3
    no, its definitely not a dirac delta function.

    In a 7-th order interpolation scheme it looks like:

    http://img855.imageshack.us/img855/3887/capturekp.png [Broken]

    by the way: i noticed this thread is probably in the wrong forum. Could someone move it to 'Atomic, Solid State, Comp. Physics', please?
    Last edited by a moderator: May 5, 2017
  5. Jun 24, 2011 #4

    Stephen Tashi

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    I don't think that is true.

    I only got a hazy grasp of those notes, but, thinking about the case in 1 dimension, the constant function f(x) = 1, when scaled as [itex] f(x)\phi(x) [/itex] is still the constant function 1. (I haven't mastered the notation. By [itex] \phi(x) [/itex], I mean a function such that when x is between the integers k and k + 1, [itex] \phi(x) [/itex] is the sum of the interpolating scaling function centered at k and the interpolating scaling function centered at k + 1.) So one can say that the integral of [itex] (1)( \phi(x)) [/itex] from k-1/2 to k+1/2 is 1.
  6. Jun 24, 2011 #5

    Stephen Tashi

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    To get my notation straight, I'll use [itex] \phi(x) [/itex] to mean a (single) scaling function centered at 0 which vanishes for x < -1/2 and x > 1/2. (I don't know where the scaling functions in those notes are supposed to vanish. This is just to test an idea.)

    If g(x) is a function that is zero except at integers, then I gather the idea of interpolating scaling functions is extend g to a continuous function f(x). Let's say that for x between the integer values k and k + 1, we compute f(x) by adding two functions. The first function [itex] f_{k,1}(x) [/itex] is g(k) times the scaling function [itex] \phi(x-k) [/itex] and the second function [itex] f_{k,2}(x) [/itex] is g(k+1) times the scaling function [itex] \phi(x - (k+1) ) [/itex].

    When you integrate f(x) over all real numbers, its integral can be regarded as the sum of integrals of these functions. Working the exercise might rely on arguing that the sum of these integrals is term-by-term equal to the sum of the non-zero values of g(x).
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